- 695

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Is it fair to say that[tex]\frac{du}{dx} = \frac 1 {dx/du}

[/tex]since this comes (I think) from the chain rule,[tex]

\frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1

[/tex]

Which means that, when integrating by substitution, I can choose to do either of[tex]

\int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx

[/tex]depending on which derivative I happen to have at hand.

(Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)

The matter came out while studying Weierstrass' substitution, where[tex]

\begin{align*}

\int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}}

= \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2

\end{align*}

[/tex]because [itex]\frac {dt}{dx}[/itex] is easier to figure out than [itex]\frac {dx}{dt}[/itex].

Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.